150+235=
300+120=
209+230=
mn chỉ em bài này với
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\(\dfrac{x+1}{2x^2-x^4}=\dfrac{x+1}{x^2\left(2-x^2\right)}=\dfrac{-\left(x+1\right)\left(x^4+2x^2+4\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\\ \dfrac{x}{x^4+2x^2+4}=\dfrac{x^3\left(x^2-2\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\\ \dfrac{2x-1}{x^7-8x}=\dfrac{2x-1}{x\left(x^6-8\right)}=\dfrac{x\left(2x-1\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\)
a, 214+[120-(214+120)]=214+[120-214-120]
=214+120-214-120
=(214-214)+(120-120)
=0+0
=0
b,(-321)-[(-321+35)-235]=-321-[-321+35-235]
=-321-321-35+235
= (-321-321)-(35+235)
=0+270
=270
what is her mother going to prepare for her bỉthdat party
2 I played it yesterday
3 I went there 2 years ago
4 I did it last night
5 I got it in June
6 I broke the......(cái j ấy) a week ago
C
1 was playing
2 was sleeping - was listening
3 was standing
4 was working - was leaving
5 Were - having
6 was running
7 was - going
8 Were watching
câu b )
ta phân phối 2 vô
=> \(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+\frac{2}{17.19}+\frac{x}{209}=\frac{10}{209}\)
\(\Rightarrow\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{17}-\frac{1}{19}+\frac{x}{209}=\frac{10}{209}\)
\(\Rightarrow\frac{8}{209}+\frac{x}{209}=\frac{10}{209}\)
\(\Rightarrow\frac{x}{209}=\frac{2}{209}\)
\(\Rightarrow x=2\)
a/ Tam giác AMN cân tại A (gt). \(\Rightarrow\) \(\widehat{AMN}=\widehat{ANM};AM=AN.\)
Xét tam giác AMB và tam giác ANC có:
+ AM = AN (cmt).
+ \(\widehat{AMB}=\widehat{ANC}\left(\widehat{AMN}=\widehat{ANM}\right).\)
+ MB = NC (gt).
\(\Rightarrow\) Tam giác AMB = Tam giác ANC (c - g - c).
\(\Rightarrow\) AB = AC (cặp cạnh tương ứng).
Xét tam giác ABC có: AB = AC (cmt).
\(\Rightarrow\) Tam giác ABC cân tại A.
b/ Tam giác ABC cân tại A (cmt) \(\Rightarrow\) \(\widehat{ABC}=\widehat{ACB}.\)
Mà \(\widehat{ABC}=\widehat{MBH;}\widehat{ACB}=\widehat{NCK}\text{}\) (đối đỉnh).
\(\Rightarrow\) \(\widehat{MBH}=\widehat{NCK}.\)
Xét tam giác MBH và tam giác NCK \(\left(\widehat{BHM}=\widehat{CKN}=90^o\right)\)có:
+ MB = NC (gt).
+ \(\widehat{MBH}=\widehat{NCK}\left(cmt\right).\)
\(\Rightarrow\) Tam giác MBH = Tam giác NCK (cạnh huyền - góc nhọn).
c/ Tam giác MBH = Tam giác NCK (cmt).
\(\Rightarrow\) \(\widehat{BMH}=\widehat{CNK}\) (cặp góc tương ứng).
Xét tam giác OMN có: \(\widehat{NMO}=\widehat{MNO}\) (do \(\widehat{BMH}=\widehat{CNK}\)).
\(\Rightarrow\) Tam giác OMN tại O.
385
420
439
HT
150+235= 385
300+120= 420
209+230 = 439
HT